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Question
AB is a diameter of the circle APBR, as shown in the figure. APQ and RBQ are straight lines. Find : ∠BPR
Solution
∠ABP = 90° – ∠BAP
= 90° – 35°
= 55°
∴ ∠ABR = ∠PBR – ∠ABP
= 115° – 55°
= 60°
∴ ∠APR = ∠ABR = 60°
(Angles subtended by the same chord on the circle are equal)
∴ ∠BPR = 90° – ∠APR
= 90° – 60°
= 30°
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